Question

(M) Use the method in this section to produce a \(QR\) factorization of the matrix in Exercise 24.

Short answer

The required factorization is, \(\left( {\begin{aligned}{{}{r}}{ - 10}&{13}&7&{ - 11}\\2&1&{ - 5}&3\\{ - 6}&3&{13}&{ - 3}\\{16}&{ - 16}&{ - 2}&5\\2&1&{ - 5}&{ - 7}\end{aligned}} \right) = \left( {\begin{aligned}{{}{r}}{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{{\sqrt 3 }}}&0\\{\frac{1}{{10}}}&{\frac{1}{2}}&0&{\frac{1}{{\sqrt 2 }}}\\{\frac{{ - 3}}{{10}}}&{ - \frac{1}{2}}&{\frac{1}{{\sqrt 3 }}}&0\\{\frac{4}{5}}&0&{\frac{1}{{\sqrt 3 }}}&0\\{\frac{1}{{10}}}&{\frac{1}{2}}&0&{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{20}&{ - 20}&{ - 10}&{10}\\0&6&{ - 8}&{ - 6}\\0&0&{6\sqrt 3 }&{ - 3\sqrt 3 }\\0&0&0&{5\sqrt 2 }\end{aligned}} \right)\).

Step-by-step solution

\(QR\) factorization of a Matrix

A matrix with order \(m \times n\) can be written as the multiplication of an upper triangular matrix \(R\) and a matrix \(Q\) which is formed by applying the Gram–Schmidt orthogonalization processto the \({\rm{col}}\left( A \right)\).

The matrix \(R\) can be found by the formula \({Q^T}A = R\).

Finding the matrix \(R\)

Given that, \(A = \left( {\begin{aligned}{{}{r}}{ - 10}&{13}&7&{ - 11}\\2&1&{ - 5}&3\\{ - 6}&3&{13}&{ - 3}\\{16}&{ - 16}&{ - 2}&5\\2&1&{ - 5}&{ - 7}\end{aligned}} \right)\).

Hence, enter matrix A in MATLAB.

>> A=(-10 13 7 -11; 2 1 5 3; -6 3 13 -3; 16 -16 -2 5; 2 1 -5 -7);

The required function:

function (B) = GramSchmidt_N(A)

(m,n) = size(A);

(U, jb) = rref(A);

x = length(jb);

B = zeros(m,x);

for i = 1:x

C(:,i)= A(:,(jb(i)));

end

B=C;

for i = 2:x

for j = 1:i-1

B(:,i) = B(:,i) - dot(C(:,i),B(:,j))/dot(B(:,j),B(:,j))* B(:,j)

end

end

for i=1:size(B,2)

TMP=B(:,i);

TMP=TMP./(sqrt(sum(TMP.^2)));

B(:,i)=TMP;

end

end

Find the Normalised orthogonal basis:

(B) = GramSchmidt_N(A)

\(\begin{aligned}{}B = \\\begin{aligned}{{}{r}}{ - 0.5000}&{0.5000}&{0.5774}&0\\{0.1000}&{0.5000}&0&{0.7071}\\{ - 0.3000}&{ - 0.5000}&{0.5774}&0\\{0.8000}&0&{0.5774}&0\\{0.1000}&{0.5000}&0&{ - 0.7071}\end{aligned}\end{aligned}\)

Therefore, the orthogonal matrix \(Q\) is\(Q = \left( {\begin{aligned}{{}{r}}{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{{\sqrt 3 }}}&0\\{\frac{1}{{10}}}&{\frac{1}{2}}&0&{\frac{1}{{\sqrt 2 }}}\\{\frac{{ - 3}}{{10}}}&{ - \frac{1}{2}}&{\frac{1}{{\sqrt 3 }}}&0\\{\frac{4}{5}}&0&{\frac{1}{{\sqrt 3 }}}&0\\{\frac{1}{{10}}}&{\frac{1}{2}}&0&{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right)\).

Find the matrix \(R\) using the following formula:

\(R = {Q^T}A\)

R=Q'*A

\(\begin{aligned}{}R = \\\begin{aligned}{{}{}}{20.0000}&{ - 20.0000}&{ - 10.0000}&{10.0000}\\0&{6.00000}&{ - 8.0000}&{ - 6.0000}\\0&0&{10.3923}&{ - 5.1962}\\0&0&0&{7.0711}\end{aligned}\end{aligned}\)

Hence, the matrix is\(R = \left( {\begin{aligned}{{}{}}{20}&{ - 20}&{ - 10}&{10}\\0&6&{ - 8}&{ - 6}\\0&0&{6\sqrt 3 }&{ - 3\sqrt 3 }\\0&0&0&{5\sqrt 2 }\end{aligned}} \right)\).